Timeline for Groups with cyclic radicals
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2018 at 7:06 | comment | added | Michal Ferov | @YCor you are spot on I added alternative definition as well. The original motivation was to study separability of cyclic subgroups in pro-$p$ topology. By necessity, if $H < G$ is $p$-closed in $G$ then it must be $p$-isolated. Informally, $H$ is closed with respect to taking $q$th roots, where $q$ is a prime distinct from $p$: $f^q \in H$ implies $f \in H$. The idea was to come up with a notion that allows me to give a nice characterisation of cyclic subgroups with such property. In this case, I can show that $\langle g \rangle$ is $p$-isolated iff $plog_G(g)$ is a power of $p$. | |
| Jun 13, 2018 at 6:56 | history | edited | Michal Ferov | CC BY-SA 4.0 |
Added further info
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| Jun 13, 2018 at 6:30 | comment | added | Michal Ferov | @MarkSapir thank you for poiting these out. As a common MO user, I'm not too familiar with Tarski mosters, I'll give it a look. | |
| Jun 9, 2018 at 10:32 | comment | added | YCor | It seems that every finitely generated torsion-free linear group in characteristic zero has a finite index subgroup satisfying (a). This is (would be?) an improvement of Selberg's lemma, and I think it can be proved with a little $p$-adic analysis. In particular, $SL_3(\mathbf{Z})$ has a finite index subgroup satisfying your condition. | |
| Jun 9, 2018 at 7:52 | comment | added | YCor | By the way, your property is equivalent to a combination of two properties, which are worth being treated separately: (a) that $Rad_G(g)$ is locally cyclic for every $g\neq 1$. This means that any two elements with a common power have a common root. (b) every locally cyclic subgroup is cyclic. For instance, any torsion -free subgroup of $SL_2$ satisfies (a). And (a) is a 1st-order invariant, since it's equivalent to the condition that any two elements with a common power commute. Hence it's equivalent to a combination of infinitely many 1st-order formulas, and passes to inductive limits. | |
| Jun 9, 2018 at 4:23 | comment | added | user6976 | As is usual in MO, you missed the torsion-free Tarski monsters (and many other groups where the centralizer of every non-identity element is cyclic). You also missed the R. Thompson group $F$. | |
| Jun 9, 2018 at 2:53 | comment | added | Michal Ferov | Actually, the property of having "cyclic radicals" is preserved w.r.t. to graph products and amalgamation along a common retract, but that's not easy to see. I am using this property to study certain behaviour and I was wondering which groups do have it, on top of the obvious ones. | |
| Jun 9, 2018 at 1:55 | comment | added | YCor | Of course you have finite direct products of such groups. Other examples are torsion-free finitely generated subgroups in $\mathrm{SL}_2(A)$, where $A$ is is the ring of integral elements in $\mathbf{Q}$. For instance, this includes torsion-free finite-index subgroups in $SL_2(\mathbf{Z}[\sqrt{2}])$, which are far from relatively hyperbolic and are not virtually residually torsion-free nilpotent. | |
| Jun 9, 2018 at 0:32 | history | edited | Michal Ferov | CC BY-SA 4.0 |
Fixed a mistake, as pointed out by Yves
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| Jun 9, 2018 at 0:28 | comment | added | YCor | "Clearly a subgroup": no, it's not a subgroup in general. For instance in the fundamental group of the Klein bottle $\langle x,y|x^2=y^2\rangle$, both $x$ and $y$ belong to $Rad_G(x)$, but $xy$ does not. Stlll the question makes sense (you're asking when is the subset $Rad_G(g)$ a cyclic subgroup for every $g\neq 1$. | |
| Jun 9, 2018 at 0:19 | review | First posts | |||
| Jun 9, 2018 at 0:45 | |||||
| Jun 9, 2018 at 0:19 | history | asked | Michal Ferov | CC BY-SA 4.0 |